 Solution

The zero-bias gate period in the first plot is $\Delta V_g=20\mathrm{mV}$. In the constant interaction model, changing the gate-induced island charge by one electron givesThe tip is at $|V_{sd}|=5.0\mathrm{mV}$ above or below zero; use this half-height, not the full $10\mathrm{mV}$ vertical extent. The addition energy is $E_{\mathrm{add}}=e(5.0\mathrm{mV})=e^2/C_{\Sigma }$, giving

To distinguish the junction capacitances, explicitly adopt the circuit convention $V_s=V_{sd}$ and $V_d=0$. The diamond with zero-bias endpoints $20$ and $40\mathrm{mV}$ has its positive tip near $(V_g,V_{sd})=(35,5)\mathrm{mV}$. Its two edge slopes are therefore $+1/3$ and $-1$.

Writing the island electrochemical potential as a charge-dependent constant minus $e(C_s{V}_{sd}+C_g{V}_g)/C_{\Sigma }$, alignment with the drain gives slope $-C_g/C_s$, while alignment with the source gives $C_g/(C_{\Sigma }-C_s)$. ThusThe source/drain names are exchanged if the opposite lead is biased under the voltage convention. The figure alone establishes the pair $\{8,16\}\mathrm{aF}$ unless that wiring convention is specified. Symmetric-looking diamond heights do not imply equal junction capacitances.